I'm trying to formalize the step in the next image from Wikipedia's article about Euler's proof of Basel problem (https://en.wikipedia.org/wiki/Basel_problem#Euler's_approach)
I've proven by induction that if we define $P_n(z)=\prod_{k=1}^n \left(1-\frac{z^2}{\pi^2}\right)$ then it is possible to express $$P_n(z)=1- \dfrac{1}{\pi^2}\left( \sum_{k=1}^n \dfrac{1}{k^2} \right)z^2+ \sum_{k=2}^n a_{k,n} z^{2k}$$ for some real numbers $a_{k,n}$.
Then, isolating the last sum and taking limits we get that
$$\dfrac{\sin(z)}{z}-1+\dfrac{1}{\pi^2}\left( \sum_{k=1}^\infty \dfrac{1}{k^2} \right)z^2=\lim_{n \rightarrow +\infty} \sum_{k=2}^n a_{k,n} z^{2k}$$
So, I know that the last sum converges and I've tried to prove that $a_{k,n}$ converges to some $a_k$ for every $k\geq 2$ and that \begin{align}{ \lim_{n \rightarrow +\infty} \sum_{k=2}^n a_{k,n} z^{2k}=\sum_{k=2}^\infty a_{k} z^{2k} \hspace{10mm}(1) }\end{align}
because if a I get that result, then
$$\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{2k}=\dfrac{\sin(z)}{z}=1-\dfrac{1}{\pi^2}\left( \sum_{k=1}^\infty \dfrac{1}{k^2} \right)z^2 + \sum_{k=2}^\infty a_{k} z^{2k}$$
and by the uniqueness of power series expansion, we can now compare the desired coefficients.
Any ideas on how to prove the convergence of the $a_{k,n}$ and (1)?
Or maybe, if there is an easier way of formaly proving we can compare the coefficients.
Thanks a lot.
I think I've thought another way of solving it.
As it can be shown by Weierstrass' factorization theorem and all that complex analysis machinery, $P_n$ converges uniformly on compact subsets of $\mathbb{C}$, so as we have a sequence of analytic functions that converges uniformly on compact subsets, its derivatives also converges uniformly on compact subsets.
So, as we have that
$$P''_n(z)=-\dfrac{2}{\pi^2}\sum_{k=1}^n \dfrac{1}{k^2}+(2k)(2k-1)\sum_{k=2}^n a_{k,n} z^{2(k-1)}$$
then
$$P''_n(0)=-\dfrac{2}{\pi^2}\sum_{k=1}^n \dfrac{1}{k^2}$$
and so by means of the convergence we have ensure before, if we denote $f(z)=\frac{\sin(z)}{z}$ then
$$f''(0)=-\dfrac{2}{\pi^2}\sum_{k=1}^\infty \dfrac{1}{k^2}$$
that was ultimately what we wanted to compare.